Question

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A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 79.8 N at an angle of 28.0° above the horizontal. The box has a mass of 20.0 kg, and the coefficient of kinetic friction between box and floor is 0.300. (Indicate the direction with the sign of your answer.) (a) Find the acceleration of the box. (Assume that the +x-axis is to the right and the ty-axis is up along the page.) magnitude 1.15 counterclockwise from the +x-axis (b) The student now starts moving the box up a 10.0° incline, keeping her 79.8 N force directed at 28.0° above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box? (Assume that the +x-axis is along the incline and the +y-axis is upwards and perpendicular to the incline.) magnitude5 direction m/s2 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. counterclockwise from the incline Need Help? Read It

Answer 1

sigma F_y = 0 f sin 28 degree + n = mg cos 10 degree n = mg cos 10 degree - F sin 28 degree m = 20 kg mu_k = 0.300 f = 79.8 N Apply Newtons's second law sigma F_x = ma F cos 28 degree - mg sin 10 degree - fk = ma F cos 28 degree -mg sin 10 degree - mu_k^n = ma F cos 28 degree - mg sin 10 degree - mu_k[mg cos 10 degree - F sin 28 degree] = ma F[cos 28 degree + mu_k sin 28 degree] - mg[sin 10 degree + mu_k cos 10 degree] = 20 a 81.7 - 91.9 = 20a a = -0.51 m/s62 a = -0.5 m/s^2 magnitude a = 0.5 m/s^2 Based on this acceleration of a body on a in circle is down wards a makes an angle 180 degree with + ve x - axis in counter clock wise theta = 180 degree